How to control Wasserstein distance in terms of characteristic function

Question

Let $\mathcal P(\Omega)$ be the set of probability measures supported on some compact subset $\Omega\subset\mathbb R^d$. For $\mu\in\mathcal P(\Omega)$, denote by $F_{\mu}$ its characteristic function, i.e.

$$F_{\mu}(x)~:=~\int_{\mathbb R^d}e^{i\langle x,y\rangle}\mu(dy)~=~\int_{\Omega}e^{i\langle x,y\rangle}\mu(dy),\quad \forall x\in\mathbb R^d.$$

Let $W(\cdot,\cdot)$ be the Wasserstein distance of order $1$. My question is whether there exists a continuous function $c:\mathbb R_+\to\mathbb R_+$ with $c(0)=0$ s.t. it holds for all $\mu$, $\nu\in\mathcal P(\Omega)$:

$$W(\mu,\nu)~\le~c\big(\|F_{\mu}-F_{\nu}\|\big),\quad \mbox{with } \|F_{\mu}-F_{\nu}\|:=\max_{x\in\mathbb R^d}|F_{\mu}(x)-F_{\nu}(x)|.$$

Any answer, comments and references are highly appreciated!

Personal thought: We use Kantorovich's duality, i.e.

$$W(\mu,\nu)~=~ \sup_{f\in \rm{Lip}_1(\mathbb R^d)} \int f d\mu -\int fd\nu ~=~\sup_{f\in \rm{Lip}_1(\Omega)} \int f d\mu -\int fd\nu,$$

where $\rm{Lip}_1(\mathbb R^d)$ and $\rm{Lip}_1(\Omega)$ denote respectively the collection of $1-$Lipschitz functions on $\mathbb R^d$ and $\Omega$. It remains to write the integral $\int f d\mu$ in terms of $F_{\mu}$. Denote by $\hat f$ the Fourier transform of $f$, i.e.

$$\hat{f}(x)~:=~ \int_{\mathbb R^d}e^{i\langle x,y\rangle}f(y)dy.$$

We notice $\hat{f}$ is well defined for each $f\in \rm{Lip}_1(\Omega)$. Then we may apply Parseval's equality under "suitable conditions"

$$\int_{\mathbb R^d}f(x)\mu(dx)~=~\int_{\mathbb R^d}\hat{f}(x) F_{\mu}(x)dx.~~~~~~~~~~~~ (\ast)$$

I am not familiar with the regularity analysis of $\hat f$. If someone knows how to proceed based on $(\ast)$, I am happy to know.

Answer 1

For measures on the real line, There is no such function $c$. Given $n$, let $\mu_n$ assign mass $1/n$ to each of $-n,n$ and remaining mass to zero. Let $\nu_n$ be Dirac measure at zero. As $n$ grows, the Fourier transforms get uniformly close, with sup norm distance of order $1/n$, yet the Wasserstein distance remains 2.

For measures on a fixed compact set there is such a function $c $ since both metrics metrize the weak * topology and the space of probability measures on a compact set is itself weak * compact.

Answer 2

You might want to look for the concept called "Fourier-based metrics/distances", also known as Toscani distances, developed by Giuseppe Toscani and his colleagues.